Let `g(x)=2f(x/2)+f(1-x)` and `f^(primeprime)(x)<0` in `0<=x<=1` then g(x)

Options

A. decreases in `[0,(2)/(3)]`
B. decreases `((2)/(3),1]`
C. increases in `[0,(2)/(3))`
D. increases in `[(2)/(3),1]`

Answer

To solve the problem, we need to analyze the function \( g(x) = 2f\left(\frac{x}{2}\right) + f(1-x) \) and determine the intervals where it is increasing or decreasing, given that \( f''(x) < 0 \) for \( 0 \leq x \leq 1 \).

### Step-by-Step Solution:

1. **Differentiate \( g(x) \)**:
\[ g'(x) = \frac{d}{dx}\left(2f\left(\frac{x}{2}\right)\right) + \frac{d}{dx}\left(f(1-x)\right) \]
Using the chain rule:
\[ g'(x) = 2 \cdot f'\left(\frac{x}{2}\right) \cdot \frac{1}{2} + f'(1-x) \cdot (-1) \]
Simplifying gives:
\[ g'(x) = f'\left(\frac{x}{2}\right) - f'(1-x) \]

2. **Determine when \( g'(x) > 0 \) (increasing)**:
\[ g'(x) > 0 \implies f'\left(\frac{x}{2}\right) > f'(1-x) \]
Since \( f''(x) < 0 \), \( f'(x) \) is a decreasing function.

3. **Analyze the inequality**:
Since \( f' \) is decreasing, if \( \frac{x}{2} < 1 - x \), then:
\[ f'\left(\frac{x}{2}\right) > f'(1-x) \]
This gives us:
\[ \frac{x}{2} < 1 - x \implies x < 2 - 2x \implies 3x < 2 \implies x < \frac{2}{3} \]

4. **Determine the interval for increasing \( g(x) \)**:
Thus, \( g(x) \) is increasing for:
\[ 0 \leq x < \frac{2}{3} \]

5. **Determine when \( g'(x) < 0 \) (decreasing)**:
\[ g'(x) < 0 \implies f'\left(\frac{x}{2}\right) < f'(1-x) \]
This occurs when \( \frac{x}{2} > 1 - x \):
\[ \frac{x}{2} > 1 - x \implies x > 2 - 2x \implies 3x > 2 \implies x > \frac{2}{3} \]

6. **Determine the interval for decreasing \( g(x) \)**:
Thus, \( g(x) \) is decreasing for:
\[ \frac{2}{3} < x \leq 1 \]

### Final Conclusion:
- \( g(x) \) is **increasing** on the interval \( [0, \frac{2}{3}) \).
- \( g(x) \) is **decreasing** on the interval \( (\frac{2}{3}, 1] \).